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अध्याय 8: ठोसों के यांत्रिक गुण

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CBSE • कक्षा XI • Physics • अध्याय 8

अनुमानित समय: 45 Mins

प्रगति: अध्ययनरत

ठोसों के यांत्रिक गुण

In Class 11 Physics, "Mechanical Properties of Solids" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

अध्याय CBT ऑनलाइन टेस्ट दें सभी अध्याय देखें

प्रस्तावना अध्याय अवधारणाएँ अध्याय सारांश स्व-मूल्यांकन अभ्यास टेस्ट दें

🏗️ Have You Ever Wondered?

Why are steel girders used to construct bridges designed with an 'I'-shaped cross-section rather than solid rectangular beams, and why is steel scientifically considered far more elastic than soft stretchy rubber? Hooke's Law and Young's Modulus quantify material elasticity.

यह अध्याय क्यों महत्वपूर्ण है

In Class 11 Physics, "Mechanical Properties of Solids" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

अध्ययन से पूर्व (आवश्यक ज्ञान)

Forces and deformations from Class 9.
Stress and pressure units ($N/m^2$ or Pascal).
Potential energy curves.

इस अध्याय के लक्ष्य

Define Elasticity, Plasticity, Deforming force, and Restoring force.
Define Stress (Tensile, Compressive, Shear, Hydraulic) and Strain (Longitudinal, Shearing, Volume).
State Hooke's Law: Within elastic limit, $\text{Stress} \propto \text{Strain}$.
Define Moduli of Elasticity: Young's Modulus ($Y$), Shear Modulus ($G$), and Bulk Modulus ($B$).
Analyze the Stress-Strain Curve for a metallic wire (Proportional limit, Yield point, Tensile strength, Fracture point).
Calculate Elastic Potential Energy stored in a stretched wire: $U = \frac{1}{2} \times \text{Stress} \times \text{Strain} \times \text{Volume}$.

अध्याय रूपरेखा एवं प्रगति

1 1. Stress, Strain & Hooke's Law

2 2. The Stress-Strain Curve

3 3. Moduli of Elasticity & 'I'-Beams

सम्पूर्ण सैद्धांतिक एवं वैचारिक अध्ययन

1. Stress, Strain & Hooke's Law

When a deforming force alters a body's shape, internal restoring forces arise:
• Stress: Restoring force per unit area: $\sigma = \frac{F}{A}$ ($\text{N/m}^2$ or $\text{Pa}$).
• Strain: Fractional deformation: $\varepsilon = \frac{\Delta L}{L}$ (dimensionless!).
• Hooke's Law: Within the elastic limit, stress is directly proportional to strain: $$\mathbf{\frac{\text{Stress}}{\text{Strain}} = E \quad (\text{Modulus of Elasticity})}$$

2. The Stress-Strain Curve

Subjecting a metal wire to increasing load reveals critical mechanical boundaries:
• Proportional Limit: Linear Hooke's region ($O$ to $A$).
• Yield Point / Elastic Limit ($B$): Maximum stress where body returns to original shape upon unloading.
• Plastic Region: Permanent deformation (strain persists even after zero stress).
• Ultimate Tensile Strength ($D$): Maximum load wire can support before thinning into a neck.
• Fracture Point ($E$): Point of catastrophic structural snapping.

3. Moduli of Elasticity & 'I'-Beams

Young's Modulus: $Y = \frac{F/A}{\Delta L/L} = \frac{F L}{A \Delta L}$ (Resistance to stretching; Steel has greater $Y$ than rubber, hence steel is more elastic!).
Bulk Modulus: $B = -\frac{\Delta P}{\Delta V/V}$ (Resistance to compression; Compressibility $k = 1/B$).
Engineering 'I'-Beams: Sagging of a beam under load is $\delta = \frac{W L^3}{4 Y b d^3}$. Making depth $d$ large drastically minimizes sagging, while the 'I'-cross section saves weight and prevents buckling!

Mechanical Properties of Solids - Key Conceptual & Analytical Model

अध्याय का सार संक्षेप एवं 10 मुख्य निष्कर्ष

मुख्य बिंदु 1

Hooke's Law: Stress is proportional to strain up to the proportional limit.

मुख्य बिंदु 2

Elastic Modulus: Intrinsic material constant resisting structural distortion.

मुख्य बिंदु 3

Steel vs Rubber: Steel is scientifically more elastic because it requires vastly greater restoring stress for deformation.

मुख्य बिंदु 4

Yield Strength: Threshold beyond which permanent plastic deformation occurs.

मुख्य बिंदु 5

I-Section Girders: Structural geometry maximizing depth $d$ to eliminate bridge sagging ($W L^3 / 4 Y b d^3$).

स्व-मूल्यांकन अभ्यास (Check Your Understanding)

मूल वैचारिक स्पष्टता की जांच के लिए नैदानिक प्रश्न। पहले स्वयं हल करें, फिर उत्तर देखें।

Why is steel considered more elastic than rubber in physics?

उत्तर एवं व्याख्या देखें

उत्तर: Elasticity measures the capacity of a material to resist deformation and generate restoring stress. For a given deforming strain, steel produces a much greater restoring stress than rubber, yielding a far higher Young's Modulus ($Y_{\text{steel}} \gg Y_{\text{rubber}}$).
Steel has a much higher Young's modulus than rubber.

A copper wire of length 2.2 m and a steel wire of length 1.6 m, both of diameter 3.0 mm, are connected end to end. When stretched by a load, the net elongation is 0.70 mm. Find the load ($Y_{\text{steel}} = 2.0 \times 10^{11}\text{ Pa}, Y_{\text{copper}} = 1.1 \times 10^{11}\text{ Pa}$).

उत्तर एवं व्याख्या देखें

उत्तर: Same tension $F$ and radius $r = 1.5\text{ mm}$. $\Delta L_{\text{total}} = \Delta L_s + \Delta L_c = \frac{F L_s}{A Y_s} + \frac{F L_c}{A Y_c} = \frac{F}{\pi r^2}\left(\frac{1.6}{2\times 10^{11}} + \frac{2.2}{1.1\times 10^{11}}\right) = 0.70 \times 10^{-3}\text{ m}$. Solving: $F = 1.8 \times 10^2\text{ N} \approx 177\text{ Newtons}$.
Load F ≈ 177 N.

Define Bulk Modulus and Compressibility. What is the bulk modulus of an ideal perfectly rigid body?

उत्तर एवं व्याख्या देखें

उत्तर: Bulk Modulus ($B$) is the ratio of hydraulic stress to volumetric strain: $B = -\frac{\Delta P}{\Delta V/V}$. Compressibility is $k = 1/B$. For a perfectly rigid body, $\Delta V = 0$, so Bulk Modulus is infinite ($\infty$).
Bulk modulus is -ΔP / (ΔV/V); infinite for rigid body.

Derive the expression for elastic potential energy stored in a stretched wire.

उत्तर एवं व्याख्या देखें

उत्तर: Work done in extending wire by $dL$ under force $F = \frac{Y A L'}{L}$ is $dW = F dL'$. Total work $W = \int_0^{\Delta L} \frac{Y A L'}{L}\, dL' = \frac{Y A (\Delta L)^2}{2L} = \frac{1}{2} F \Delta L = \frac{1}{2} \times \text{Stress} \times \text{Strain} \times \text{Volume}$.
U = 1/2 × Stress × Strain × Volume.

Why are bridge girders designed with an 'I'-shaped cross section?

उत्तर एवं व्याख्या देखें

उत्तर: The depression (sagging) of a beam of length $L$, breadth $b$, and depth $d$ under central load $W$ is $\delta = \frac{W L^3}{4 Y b d^3}$. Increasing depth $d$ reduces sagging cubicly, while the 'I'-shape concentrates mass at the top and bottom flanges to resist bending without buckling.
Maximizes depth to reduce sagging by d^3 factor while saving weight.

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